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Looking closely again, I am pretty sure my commentmy comment gives the answer. Note that Hörmander uses the notation $D = - i \partial$. So in particular

$$ Du = -i (n\pi i) u $$

when $u = e^{n\pi i x}$. This gives an extra minus sign in the first term ($\nu = 1$ and $\mu = 0$) of the definition for $L_1 u$, while leaving the second term ($\nu = 2$ and $\mu = 1$) unchanged, providing the extra $(-1)^\nu$ you are looking for.

Looking closely again, I am pretty sure my comment gives the answer. Note that Hörmander uses the notation $D = - i \partial$. So in particular

$$ Du = -i (n\pi i) u $$

when $u = e^{n\pi i x}$. This gives an extra minus sign in the first term ($\nu = 1$ and $\mu = 0$) of the definition for $L_1 u$, while leaving the second term ($\nu = 2$ and $\mu = 1$) unchanged, providing the extra $(-1)^\nu$ you are looking for.

Looking closely again, I am pretty sure my comment gives the answer. Note that Hörmander uses the notation $D = - i \partial$. So in particular

$$ Du = -i (n\pi i) u $$

when $u = e^{n\pi i x}$. This gives an extra minus sign in the first term ($\nu = 1$ and $\mu = 0$) of the definition for $L_1 u$, while leaving the second term ($\nu = 2$ and $\mu = 1$) unchanged, providing the extra $(-1)^\nu$ you are looking for.

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Willie Wong
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Looking closely again, I am pretty sure my comment gives the answer. Note that Hörmander uses the notation $D = - i \partial$. So in particular

$$ Du = -i (n\pi i) u $$

when $u = e^{n\pi i x}$. This gives an extra minus sign in the first term ($\nu = 1$ and $\mu = 0$) of the definition for $L_1 u$, while leaving the second term ($\nu = 2$ and $\mu = 1$) unchanged, providing the extra $(-1)^\nu$ you are looking for.